By induction on the number of constraints $m$ employed in the definition of table $R$.

\noindent - If $m=0$, that is, $R$ includes neither primary key nor foreign key constraints.
Then, by Definition \ref{def:sqlop}, $\SQL{T,d} = d(T)=\multi{\mu_1,\dots,\mu_n}$.
By Definition \ref{def:constraints},  $\theta(T,d)$ $= \multi{ (true,\mu_1), \dots, (true,\mu_n) }$.
Then, every element $\eta$  occurs in $\SQL{R,d}$  with the same cardinality as $(true,\eta)$ in $\theta(R,d)$.


\noindent - If $m>0$, then $R$ is defined including at least one primary or foreign key constraint.
Select any constraint $C$. Suppose that $C$ is a primary key  defined by columns $C_1, \dots, C_p$.
Then, every $\mu_i \in \SQL{R,d}$, appears only once  in $\SQL{R,d}$  due to the primary key constraint.
We check that $\mu_i \in \SQL{R,d}$ iff $(true,\mu_i) \in \theta(R,d)$ (with cardinality $1$).
From Definition \ref{def:pkandfk}.\ref{def:sql:primary}, $\mu_i \in \SQL{R,d}$, with $d$ a valid instance implies that
\begin{align}
&&&\bigwedge_{j=1, j\neq i}^n(\bigvee_{k=1}^p \mu_i(R.C_k) \neq \mu_j(R.C_k) ) \label{demo:tablepk}
\end{align}
Define $T'$ as $R$ without this constraint, as explained in Definition \ref{def:constraints}.\ref{def:constraints:primary}. Then, $d$ is also a valid instance for $T'$,  and
thus  $d(T') = d(R)$ which means $\SQL{T',d} = \SQL{(R,d)}$.
Assume that $\theta(T',d) = \multileft (\psi_1,\mu_1),$ $\dots,$ $(\psi_n,\mu_n) \multiright$, and from Definition
\ref{def:constraints}.\ref{def:constraints:primary}
\begin{align}
\theta(T,d) = \multi{(\psi_i \wedge (\bigwedge_{j=1, j\neq i}^n(\bigvee_{k=1}^p \mu_i(R.C_k) \neq \mu_j(R.C_k) ))),\mu_i)|i \in 1,\dots,n} \label{demo:thetapk}
\end{align}

$\mu_i \in \SQL{R,d}$ iff (from  $\SQL{T',d} = \SQL{(R,d)}$) $\mu_i \in \SQL{T',d}$ iff $(true,\mu_i) \in \theta(T',d)$ (
by induction hypothesis)  iff $\psi_i = true$.
Moreover,  $\mu_i \in \SQL{R,d}$ iff \ref{demo:tablepk} is $\top$ (because $d$ is valid)  iff  (true,$\mu_i$) $\in \theta(R,d)$ (replacing $\psi_i$ and \ref{demo:tablepk}  by $\top$ in
\ref{demo:thetapk}).
The result is analogous if the selected constraint $C$ is a foreign key.




%If $(true,\mu_a), (true, \mu_b) \in \SQL{R,d}$ and $\mu_a = \eta, mu_b = \eta$ then from \ref{def:constraints:primary}
%
%$(\bigwedge_{j=1, j\neq a}^n(\bigvee_{k=1}^p \eta(T.C_k) \neq \mu_j(T.C_k) )) \wedge$
%
%$(\bigwedge_{j=1, j\neq b}^n(\bigvee_{k=1}^p \eta(T.C_k) \neq \mu_j(T.C_k) ))$
%
%The previous formula can not be true if $a \neq b$ so it can not be more than one $(true,\eta) \in \theta(R,d)$.



